50 gm of `CaCO_(3)` is allowed to react with 68.6 gm of `H_(3)PO_(4)` then select the correct option(s):
`3CaCO_(3) +2H_(3)PO_(4) rarr Ca_(3)(PO_(4))_(2)+3H_(2)O+3CO_(2)`

Options

A. 51.67 gm salt is formed
B. Amount of unreacted reagent =35.93 gm
C. `n_(Co_(2))` =0.5 moles
D. 0.7 moles of `CO_(2)` is evoled

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the moles of `CaCO3`
1. **Given mass of `CaCO3`** = 50 g
2. **Molar mass of `CaCO3`** = 100 g/mol
3. **Moles of `CaCO3`** = Given mass / Molar mass = 50 g / 100 g/mol = 0.5 moles

### Step 2: Calculate the moles of `H3PO4`
1. **Given mass of `H3PO4`** = 68.6 g
2. **Molar mass of `H3PO4`** = 98 g/mol
3. **Moles of `H3PO4`** = Given mass / Molar mass = 68.6 g / 98 g/mol = 0.7 moles

### Step 3: Determine the limiting reagent
1. From the balanced equation: `3CaCO3 + 2H3PO4 → Ca3(PO4)2 + 3H2O + 3CO2`
2. The stoichiometric ratio from the equation shows that 3 moles of `CaCO3` react with 2 moles of `H3PO4`.
3. Calculate the ratio of available moles:
- For `CaCO3`: 0.5 moles / 3 = 0.1667
- For `H3PO4`: 0.7 moles / 2 = 0.35
4. Since 0.1667 < 0.35, `CaCO3` is the limiting reagent.

### Step 4: Calculate the mass of `Ca3(PO4)2` produced
1. From the balanced equation, 3 moles of `CaCO3` produce 1 mole of `Ca3(PO4)2`.
2. Moles of `Ca3(PO4)2` produced = Moles of `CaCO3` used / 3 = 0.5 moles / 3 = 0.1667 moles
3. **Molar mass of `Ca3(PO4)2`** = (3 × 40) + (2 × (31 + 16×4)) = 310 g/mol
4. **Mass of `Ca3(PO4)2` produced** = Moles × Molar mass = 0.1667 moles × 310 g/mol = 51.67 g

### Step 5: Calculate the unreacted `H3PO4`
1. From the balanced equation, 2 moles of `H3PO4` react with 3 moles of `CaCO3`.
2. Moles of `H3PO4` consumed = (0.5 moles of `CaCO3` × 2 moles of `H3PO4`) / 3 = 0.3333 moles
3. Unreacted moles of `H3PO4` = Initial moles - Consumed moles = 0.7 moles - 0.3333 moles = 0.3667 moles
4. **Mass of unreacted `H3PO4`** = Unreacted moles × Molar mass = 0.3667 moles × 98 g/mol = 35.93 g

### Step 6: Calculate the number of moles of `CO2` produced
1. From the balanced equation, 3 moles of `CaCO3` produce 3 moles of `CO2`.
2. Moles of `CO2` produced = Moles of `CaCO3` used = 0.5 moles

### Summary of Results
- Mass of `Ca3(PO4)2` produced = 51.67 g
- Mass of unreacted `H3PO4` = 35.93 g
- Moles of `CO2` produced = 0.5 moles

### Final Options
- A: 51.67 g of `Ca3(PO4)2` (Correct)
- B: 35.93 g of unreacted `H3PO4` (Correct)
- C: 0.5 moles of `CO2` produced (Correct)
- D: 0.7 moles of `CO2` evolved (Incorrect)